Question

$2\text{g}$ of gas $A$ is introduced in an evacuated flask at ${25}^{\circ }$C. The pressure of the gas is $1$ atm. Now $3\text{g}$ of another gas $B$ is introduced in the same flask and the total pressure becomes $1.5$ atm. The ratio of molecular mass of $A$ and $B$ is:

• A. $\frac{3}{1}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

Initially 2 $g$ of gas A was introduced in the evacuated flask, its pressure being 1 $atm$.

Partial pressure of gas A = $p_A$

Let the molecular mass of the gas A be $m_A$

Moles of A, $n_A=\frac{2}{m_A}$

When $3g$ of gas B was introduced, the pressure inside the flask increases to $1.5atm$.

Partial pressure of gas B, $p_B=\text{Total pressure}-\text{Partial pressure of gas A}$
$p_B=1.5-1$
$p_B=0.5atm$

Let the molecular mass of the gas B be $m_B$

Moles of B, $n_B=\frac{3}{m_B}$

Assuming ideal gas behaviour,
At constant temperature and volume, $p\propto n$

$\therefore \frac{p_A}{p_B}=\frac{n_A}{n_B}$
Where $p$ is the partial pressure of the gas
$\frac{1}{0.5}=\frac{\frac{2}{m_A}}{\frac{3}{m_B}}$

$\frac{m_A}{m_B}=\frac{1}{3}$